mathmath333
  • mathmath333
functions
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &\normalsize \text{Find the maximim value of the function}\hspace{.33em}\\~\\ & \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)
SolomonZelman
  • SolomonZelman
interval given ?
SolomonZelman
  • SolomonZelman
with out interval this maximum value diverges to infinity.

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mathmath333
  • mathmath333
this question means that i have to minimize \(\large \color{black}{\begin{align} x^2-3x+2 \hspace{.33em}\\~\\ \end{align}}\) right ?
SolomonZelman
  • SolomonZelman
should be, or els... i mean there is not maximum value here, if you are taking over \((-\infty,~+\infty)\)
SolomonZelman
  • SolomonZelman
wait
SolomonZelman
  • SolomonZelman
1/that ? or, just that ?
mathmath333
  • mathmath333
lol its \(\large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\) and you want to find absolute minimum, right?
SolomonZelman
  • SolomonZelman
well, I can tell you that the limit as x approaches \(\pm\)infinity is going to be 0.
SolomonZelman
  • SolomonZelman
we can plot, I am guessing it is something like |dw:1435010239497:dw|
mathmath333
  • mathmath333
i want to find the maximum value of this \(\large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)
SolomonZelman
  • SolomonZelman
none
SolomonZelman
  • SolomonZelman
it is +infinity
mathmath333
  • mathmath333
but how ?
SolomonZelman
  • SolomonZelman
https://www.desmos.com/calculator
mathmath333
  • mathmath333
how i prove that with algebra
SolomonZelman
  • SolomonZelman
even the minimum doesn't exist.
SolomonZelman
  • SolomonZelman
with algebra, even with no calc ?
mathmath333
  • mathmath333
idk calculus
SolomonZelman
  • SolomonZelman
ic
SolomonZelman
  • SolomonZelman
lets c
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\) it has vertical asymptotes, can you find them for me?
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} x=2,\ 1 \hspace{.33em}\\~\\ \end{align}}\)
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
and, the best thing I can see here is to use the idea of a limit to see where range tends to the close we get to x=1 and x=2. I don't really know a purely algebraic proof for why this function has no max or min.
xapproachesinfinity
  • xapproachesinfinity
that has a max actually
SolomonZelman
  • SolomonZelman
yes?
SolomonZelman
  • SolomonZelman
https://www.desmos.com/calculator keep scrollin up
SolomonZelman
  • SolomonZelman
on a certain interval, no doubt (unless it has a domain gap)
xapproachesinfinity
  • xapproachesinfinity
i think it attains one max does not perhaps required any domain to get it or i think
xapproachesinfinity
  • xapproachesinfinity
your graph does not show up see here https://www.desmos.com/calculator
SolomonZelman
  • SolomonZelman
i don't think there is a max. that is due to an asymptote - two of them, there.
SolomonZelman
  • SolomonZelman
we can find the limit from the right and from the left as x approaches 1 and 2.
xapproachesinfinity
  • xapproachesinfinity
http://prntscr.com/7k4q1l here see that bottom part of the graph
xapproachesinfinity
  • xapproachesinfinity
with calculus we can prove it has a max value but @mathmath333 is looking for a different way
SolomonZelman
  • SolomonZelman
this graph goes further down and up if you scroll on desmos, but we can go ahead and take any limit that is closed from either side of any of the asymptotes you wish to choose.....
xapproachesinfinity
  • xapproachesinfinity
i mean here local max not abolute of course
mathmath333
  • mathmath333
how to find that local max
SolomonZelman
  • SolomonZelman
oh, but the question is asking for absolute doesn't it?
SolomonZelman
  • SolomonZelman
that is what i would think of course
xapproachesinfinity
  • xapproachesinfinity
first let's write that as 1/(x-1)(x-2)
xapproachesinfinity
  • xapproachesinfinity
i had thought of absolute too, just not sure about it!
mathmath333
  • mathmath333
yes the question is asking absolute, i just asked that of curiosity
SolomonZelman
  • SolomonZelman
this function is pretty weird. I wonder what would it model
SolomonZelman
  • SolomonZelman
i mean in real world
xapproachesinfinity
  • xapproachesinfinity
you want to then prove there is no max algebraically
mathmath333
  • mathmath333
yes ^
SolomonZelman
  • SolomonZelman
would that include using ideas from calc but not its techniques?
SolomonZelman
  • SolomonZelman
I am stubborn on the idea of a limit. A limit would simply show that as it approaches __ it can be infinitely large/small
mathmath333
  • mathmath333
ok i will try to bear with calculus
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\) we need to do only 1 asymptote for this
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\) lets plug in values that are less than, but close to 1
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f(0.9)=? }\)
SolomonZelman
  • SolomonZelman
= about 9
SolomonZelman
  • SolomonZelman
f(0.99)=99
SolomonZelman
  • SolomonZelman
try plugging it yourself for close values to 1
SolomonZelman
  • SolomonZelman
these values are called approaching "from the left" because they are a buit less than 1
SolomonZelman
  • SolomonZelman
now, try to do f(1.5) f(1.1) f(1.01) f(1.001) and you will see where it goes "from the right" - for values that are greater than 1
SolomonZelman
  • SolomonZelman
i got to go
SolomonZelman
  • SolomonZelman
sorry
xapproachesinfinity
  • xapproachesinfinity
suppose there exist a max max(1/(x^2-3x+2)=c 1/x^2-3x+21/c=l x^2-3x+2-l>0
xapproachesinfinity
  • xapproachesinfinity
just some random stuff lol
xapproachesinfinity
  • xapproachesinfinity
coming back..
mathmath333
  • mathmath333
ok i need to go please comment if u found something
xapproachesinfinity
  • xapproachesinfinity
let's first consider the domain \(D_f: \{ x\in \mathbb{R}: x\ne1, x\ne 2 \}\)
xapproachesinfinity
  • xapproachesinfinity
that is (-oo, 1)u(1, 2)u(2,oo)
xapproachesinfinity
  • xapproachesinfinity
we look at the sign of y when x<1 if y does not change the sign in that interval then i must not have a max seems that 1/(x-2)(x-1)>0 for x<1 same reasoning for x>2 ===> y>0 so only btw 1 and 2 left
xapproachesinfinity
  • xapproachesinfinity
from the graph we knew for 1

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